Case Study 1 — How Much Drug Did the Patient Actually Get?

Field: Clinical pharmacology / pharmacokinetics Techniques used: $u$-substitution (Section 15.1), integration by parts (Section 15.5), the net-change reading of FTC (Chapter 14).

The Setting

Priya Nair is a clinical pharmacologist running a bioequivalence study for a generic antibiotic. The question her team must answer sounds simple but is the central quantity of her entire field: over the hours after a dose, how much drug did the patient's bloodstream actually receive? Not the milligrams swallowed — the milligrams that reached the blood, integrated across time. That number is the area under the concentration–time curve, the AUC, and the regulator will not approve the generic unless its AUC matches the brand-name drug within tight bounds.

A blood draw gives concentration $C(t)$ at scattered time points. A pharmacokinetic model fits a smooth curve to those points, and then the AUC is an integral. The shape of that integral — and which Chapter 15 technique unlocks it — depends on the model. Priya works through three increasingly realistic models in an afternoon, and each one is a worked exercise in this chapter's tools.

Model 1 — Simple Exponential Decay (a $u$-Substitution)

For a drug given intravenously, concentration peaks immediately and then decays exponentially as the body clears it:

$$C(t) = C_0\, e^{-kt},$$

where $C_0$ is the initial concentration (here $C_0 = 8\ \text{mg/L}$) and $k$ is the elimination rate constant ($k = 0.25\ \text{hr}^{-1}$). The total exposure over all time is

$$\text{AUC} = \int_0^\infty C_0\, e^{-kt}\, dt.$$

This is the simplest substitution in the book. Let $u = -kt$, so $du = -k\,dt$ and $dt = -\tfrac{1}{k}\,du$:

$$\int C_0 e^{-kt}\,dt = -\frac{C_0}{k}\int e^{u}\,du = -\frac{C_0}{k}e^{-kt} + \text{const}.$$

Evaluating from $0$ to $\infty$ — the upper limit is a limit process, the improper integral we treat carefully in Chapter 17, but here the antiderivative simply decays to zero:

$$\text{AUC} = \left[-\frac{C_0}{k}e^{-kt}\right]_0^\infty = 0 - \left(-\frac{C_0}{k}\right) = \frac{C_0}{k}.$$

With Priya's numbers, $\text{AUC} = 8 / 0.25 = 32\ \text{mg·hr/L}$. The result is almost insultingly clean: total exposure is the initial concentration divided by the clearance rate. A drug that clears twice as fast (larger $k$) delivers half the total exposure, which is exactly why kidney impairment — which lowers $k$ — raises a drug's AUC and forces a dose reduction.

Connection to FTC (Chapter 14). The AUC integral is the net-change reading of the Fundamental Theorem applied to a decaying rate: the accumulated concentration-time is the antiderivative evaluated at the endpoints. Chapter 15 supplies the antiderivative that Chapter 14 told us to find.

Model 2 — Oral Dose with Absorption (the Bateman Function)

An IV dose is instantaneous, but a swallowed pill must first be absorbed before it is eliminated. The two competing exponentials give the Bateman function:

$$C(t) = \frac{F D\, k_a}{V (k_a - k)}\left(e^{-kt} - e^{-k_a t}\right),$$

where $k_a$ is the absorption rate and the leading constant bundles dose $D$, bioavailability $F$, and volume $V$. The AUC is

$$\text{AUC} = \frac{F D\, k_a}{V(k_a - k)}\int_0^\infty \left(e^{-kt} - e^{-k_a t}\right) dt.$$

Each term integrates by the same substitution as Model 1. We found $\int_0^\infty e^{-kt}\,dt = \tfrac{1}{k}$; likewise $\int_0^\infty e^{-k_a t}\,dt = \tfrac{1}{k_a}$. So

$$\text{AUC} = \frac{F D\, k_a}{V(k_a - k)}\left(\frac{1}{k} - \frac{1}{k_a}\right) = \frac{F D\, k_a}{V(k_a - k)}\cdot\frac{k_a - k}{k\,k_a} = \frac{F D}{V k}.$$

The absorption rate $k_a$ cancels completely. This is a genuinely useful surprise: the total exposure of an oral dose does not depend on how fast it is absorbed, only on how much is bioavailable ($FD$) and how fast it clears ($Vk$). Two formulations of the same drug — a fast-dissolving tablet and a slow-release capsule — can deliver identical AUC even though their concentration curves look nothing alike. That fact is the mathematical foundation of bioequivalence testing, and it fell out of two routine substitutions.

Model 3 — When the Rate Itself Carries a Polynomial (Integration by Parts)

Sometimes a drug's input rate is modeled with a polynomial factor in front of the exponential — for instance an infusion that ramps up linearly before tapering, giving an effect-site rate proportional to $t\,e^{-kt}$. Priya needs the mean residence time-style quantity

$$M = \int_0^\infty t\, e^{-kt}\, dt.$$

Now substitution alone fails: there is a stray $t$ multiplying the exponential, a product of different types of functions — the signature of integration by parts. Take $u = t$, $dv = e^{-kt}\,dt$, so $du = dt$ and $v = -\tfrac{1}{k}e^{-kt}$:

$$M = \left[-\frac{t}{k}e^{-kt}\right]_0^\infty + \frac{1}{k}\int_0^\infty e^{-kt}\,dt.$$

The boundary term vanishes at both ends: at $t=0$ the factor $t$ is zero, and as $t\to\infty$ the exponential decay outruns the linear growth, so $t\,e^{-kt}\to 0$. (Exponential decay beats any polynomial — a fact Priya will see again in every Laplace transform, Section 15.10.) The leftover integral is Model 1 again:

$$M = 0 + \frac{1}{k}\cdot\frac{1}{k} = \frac{1}{k^2}.$$

For $k = 0.25\ \text{hr}^{-1}$ this gives $M = 1/0.0625 = 16\ \text{hr}^2\cdot(\text{rate units})$ — and the ratio $M/\text{AUC}$ is exactly the mean residence time $1/k = 4$ hours, the average time a drug molecule spends in the body. Integration by parts converted a clinically meaningful average into one short calculation, and the same $u = t$, $dv = e^{-kt}\,dt$ choice powers the entire pharmacokinetic literature on moments.

What Priya Concludes

By dinner Priya has the three numbers her report needs, each from a Chapter 15 technique:

• AUC for the IV reference: a substitution, $C_0/k$.
• AUC for the oral generic: two substitutions, $FD/(Vk)$ — provably independent of absorption rate.
• Mean residence time: integration by parts, $1/k$.

Her bioequivalence verdict reduces to comparing $FD/(Vk)$ for the generic against $C_0/k$ scaled for the reference. The regulator cares about a ratio, and every piece of that ratio came from an antiderivative that a CAS could have produced in milliseconds — but Priya needed to know which closed forms to expect to design the study and read its output, exactly the judgment Section 15.7 argues hand technique buys you.

Discussion Questions

1. In Model 1, what happens to the AUC if the patient's kidneys are impaired so that $k$ drops by half? Interpret the clinical danger in one sentence.
2. Model 2 showed the absorption rate $k_a$ cancels out of the AUC. Sketch (qualitatively) two concentration curves with the same AUC but very different peak heights, and explain why peak concentration — not just AUC — also matters clinically.
3. Re-derive the boundary term in Model 3 carefully: why is it legitimate to write $\big[-\tfrac{t}{k}e^{-kt}\big]_0^\infty = 0$? Which limit is the "interesting" one, and which growth rate wins?
4. Suppose the input rate were $t^2 e^{-kt}$ instead of $t\,e^{-kt}$. How many integration-by-parts passes would you need, and what role does the tabular method (Section 15.5) play?
5. Model 3's integral is improper (its upper limit is $\infty$). Identify exactly where the improperness lives and preview how Chapter 17 will make the $\big[\cdots\big]_0^\infty$ step rigorous.

A Short Annotated Reading

• Rowland & Tozer, Clinical Pharmacokinetics and Pharmacodynamics (4th ed.). The standard text; Chapters 2–3 derive AUC and the Bateman function with exactly the integrals above. Read it to see how $C_0/k$ and $FD/(Vk)$ govern real dosing.
• Stewart, Calculus: Early Transcendentals (9th ed.), §7.1. Integration by parts, including the $\int t\,e^{-kt}\,dt$ archetype that drives Model 3.
• OpenStax, Calculus Volume 2, §3.1. Integration by parts with the same polynomial-times-exponential examples, freely available — a good companion if you want a second derivation of $M = 1/k^2$.

Return to: Chapter 15 · Next: Case Study 2